(i) If f(x)=x then f(y)=y and yf(x)=xy;
x(f(y))=xf(y) because x is not a function, but the independent variable;
f(x(f(y)))=f(xf(y))=xf(y)=xy. Hence f(x(f(y)))=yf(x)=xy when f(x)=x.
But as x→∞, f(x)→∞, therefore f(x)≠x.
(ii) Now let f(x)=1/x then f(y)=1/y; xf(y)=x/y; yf(x)=y/x; f(xf(y))=1/(xf(y))=(1/x)(1/f(y))=y/x.
As x→∞, f(x)→0. Therefore when f(x)=1/x, f(x(f(y)))=f(xf(y))=xf(y).
Hence both conditions are satisfied when f(x)=1/x.
(iii) If f(x)=a/x then f(y)=a/y and xf(y)=ax/y;
yf(x)=ay/x; f(xf(y))=y/(ax); for f(xf(y))=yf(x), a(y/x)=(1/a)(y/x); a2=1, a=±1.
Therefore for x,y>0, f(x)=1/x or -1/x.